Energy and Simple Harmonic Motion of spring

A 2.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.200m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distance above the point of release, where h = 0.

h (meters) KE PE(gravity) PE(elastic) E
0
0.200
0.400

2. Relevant equations

Ef = E0 (Conservation of mechanical energy)

K = .5mv2 Us = .5kx2 Ug = mgh

3. The attempt at a solution

I got all of the gravitational potential energies ( 0 , 3.92J , 7.84J) using Ug = mgh and KE at 0, where v = 0. For the elastic potential energy at 0, I thought that you would be able to use .5kx2 with x = 0.200m, but that doesn't give the right answer (which is 8.76 J). I think the answer has something to do with the equation for conservation of mechanical energy, but what I tried didn't yield a correct answer or even make much sense. I'm probably missing something obvious, but I would appreciate some help.

A 2.00-kg object is hanging from the end of a vertical spring. The spring constant is 50.0 N/m. The object is pulled 0.200m downward and released from rest.

For the elastic potential energy at 0, I thought that you would be able to use .5kx2 with x = 0.200m,

The elastic potential energy is kΔL2/2, where ΔL is the length the spring is stretched with. The spring was stretched by the hanging object already, and stretched further by 0.2 m. Calculate the equilibrium ΔLe from mg=kΔLe, and add 0.2 m to get the initial ΔL.